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Homework 6 Solutions
Kyle Chapman
November 17, 2011
102.4
Seeking a contradiction, suppose there are only ﬁnitely many primes of the form 4
k
+ 3. Denote them by
4
k
1
+ 3
,
4
k
2
+ 3
,...,
4
k
n
+ 3. Multiply all of them together to give
N
= (4
k
1
+ 3)(4
k
2
+ 3)
...
(4
k
n
+ 3).
Expanding this out gives us a number which is odd, so either
N
= 4
k
+1 for some
k
or
N
= 4
k
+3 for some
k
. In the ﬁrst case, let
M
=
N
+ 2 and in the second case let
M
=
N
+ 4. In both cases,
M
is of the form
4
j
+ 3 for some
j
. We know
M
= Π
l
i
=1
p
i
for some collection of primes
p
i
. Further, since
M
is odd, none of
the
p
i
can be 2. This means that every
p
i
is either of the form 4
r
+ 1 or 4
r
+ 3. If they are all of the form
4
r
+ 1 then their product would also be of the form 4
r
+ 1 which it is not, so at least one of the
p
i
is of
the form 4
r
+ 3. This means that this prime divides both
N
and
M
which means it divides their diﬀerence.
Their diﬀerence, however, is either 2 or 4 neither of which has a factor of the form 4
r
+ 3. Thus we have our
contradiction.
102.6
From the construction, each of the numbers
N,N
+ 1
,...,N
+ 9 is of the form 11! +
k
where
k
is between
2 and 11. This means that
k
(
11!
k
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